## Measurable Spaces – Problem (47/365)

The previous post involved distribution functions over the real numbers but it’s also possible to have distribution functions over $R^n$. A problem asks to show that if we have the distribution function

$\displaystyle F(x_1, \dots, x_n) = P((-\infty, x_1] \times \dots \times (\infty, x_n])$

And a difference function

$\displaystyle \Delta_{a_i,b_i} : \mathbb{R}^n \rightarrow \mathbb{R} \\ \Delta_{a_i,b_i} F(x_1, \dots, x_n) = F(x_1, \dots, x_{i-1}, b_i, x_{i+1}, \dots) - F(x_1, \dots, x_{i-1}, a_i, x_{i+1}, \dots) \\ \text{for } a_i \le b_i$

Then show that

$\displaystyle \Delta_{a_1,b_1} \dots \Delta_{a_n,b_n} F(x_1,\dots,x_n) = P(a,b] \\ \text{where } a = (a_1, \dots, a_n), b = (b_1, \dots, b_n)$

Just to make clear the notation above (which confused me for a while), take the example where $n=2$, then

$\displaystyle \Delta_{a_2,b_2}F(x_1,x_2) = F(x_1, b_2) - F(x_2, a_2) \\ \Delta_{a_1,b_1}\Delta_{a_2,b_2}F(x_1,x_2) = [F(b_1, b_2) - F(b_1, a_2)] - [F(a_1, b_2) - F(a_1, a_2)] \\ \text{this is actually } \\ = P(x_1 \le b_1, x_2 \le b_2) + P(x_1 \le a_1, x_2 \le a_2) \\ \text{ } - P(x_1 \le b_1, x_2 \le a_2) - P(x_1 \le a_1, x_2 \le b_2) \\ = P(a_1 < x_1 \le b_1, a_2 < x_2 \le b_2) \\ = P(a,b]$

So this is true for $n=2$. I won’t do the general case.